Edexcel A-Level Chemistry Paper 1, June 2025: Question 9

11 marks · Medium difficulty · Practical Techniques and Data Analysis

Analyse titration data for the reaction between acidified potassium manganate(VII) and hydrogen peroxide in hand sanitiser to determine mass percentage and experimental uncertainties.

Practise this question

Question

Question 9 presents an analysis of hydrogen peroxide (H2O2) in hand sanitiser with density 0.82 g cm⁻³. 10.00 cm³ of sanitiser is diluted to 250.0 cm³; 25.00 cm³ portions are titrated with 0.0200 mol dm⁻³ KMnO4. Part (a) provides a results table for four titrations with two missing values (titre 3 and initial reading 4). Part (b) asks how a mean titre of 15.60 cm³ was determined. Part (c) gives the ionic redox equation 2MnO4⁻ + 5H2O2 + 6H⁺ → 2Mn²⁺ + 5O2 + 8H2O and asks for the flask colour change at the end-point. Part (d) requires calculating whether the % by mass of H2O2 is within the WHO range of 3.0% to 6.0%. Part (e) gives uncertainties of ±0.04 cm³ for the pipette and ±0.05 cm³ for each burette reading, asking for the percentage uncertainties in 25.00 cm³ and 15.55 cm³.

Mark scheme

Show the mark scheme Mark scheme for Question 9: (a) titre 3 = 15.25 cm³ and initial reading 4 = 9.35 cm³ (1 mark). (b) Mean of concordant titres 2 and 4 (1 mark). (c) Colour change from colourless to (pale) pink (2 marks). (d) 5-mark calculation: moles of MnO4⁻ = 3.12 × 10⁻⁴ mol; moles of H2O2 in 25.0 cm³ = 7.80 × 10⁻⁴ mol; moles in 250 cm³ = 7.80 × 10⁻³ mol; mass of H2O2 = 0.2652 g; mass of 10 cm³ sanitiser = 8.2 g; % by mass = 3.23%, concluding it is within WHO guidelines. (e) Pipette percentage uncertainty = (0.04/25) × 100 = 0.16%; burette percentage uncertainty = (2 × 0.05/15.55) × 100 = 0.64% (2 marks).

How to answer it

Analysis of Hydrogen Peroxide in Hand Sanitiser via Redox Titration

What This Question Tests

Core practical skills, quantitative redox chemistry, and measurement uncertainties:

  • Completing titration tables and identifying concordant titres (within ±0.10 cm³ of each other) to calculate a valid mean.
  • Self-indicating redox endpoints for potassium manganate(VII) ( MnO₄⁻ ) titrations.
  • Multi-step quantitative calculation involving dilution factors, molar stoichiometry (5:2 ratio), density ( mass = density × volume ), and mass percentages.
  • Calculating apparatus percentage uncertainties, distinguishing between single-reading instruments (pipette) and dual-reading instruments (burette).

Part (a): Completing the Titration Results Table

Completing missing burette readings and titres (1 Mark)

Titration number 1 2 3 4
Final burette reading / cm³ 15.90 24.00 15.30 25.00
Initial burette reading / cm³ 0.00 8.45 0.05 9.35
Titre / cm³ 15.90 15.55 15.25 15.65

✅ Correct Values

  • Titre 3: 15.30 − 0.05 = 15.25 cm³
  • Initial reading 4: 25.00 − 15.65 = 9.35 cm³

🧠 Exam Technique

Always record burette readings to two decimal places, where the second digit is either a 0 or a 5 (e.g., 9.35 , not 9.4 ). Both values must be correct to secure the single mark.

Mark Scheme: 1 mark for both Titre 3 correct (15.25) AND Initial reading 4 correct (9.35).

Part (b): Determining the Mean Titre

Selecting concordant values for averaging (1 Mark)

✅ Correct Answer

The value was calculated by taking the mean (or average) of the concordant titres (Titres 2 and 4).

Mean = (15.55 + 15.65) / 2 = 15.60 cm³

❌ Common Errors

  • Stating that all four titrations were averaged: (15.90 + 15.55 + 15.25 + 15.65) / 4 = 15.59 cm³ . This loses the mark.
  • Failing to mention the word concordant or omitting which specific titres were used.

💡 Key Knowledge: Concordancy Rules

Concordant titres are within ±0.10 cm³ of each other. Here, Titre 2 ( 15.55 cm³ ) and Titre 4 ( 15.65 cm³ ) differ by exactly 0.10 cm³. Titre 1 is a rough/trial titration (15.90 cm³) and Titre 3 (15.25 cm³) is non-concordant.

Mark Scheme: 1 mark for reference to the mean/average of concordant titres OR identifying titres 2 and 4. Do not award if all four titres are averaged.

Part (c): End-Point Colour Change

Potassium manganate(VII) self-indicating redox endpoint (2 Marks)

Equation: 2MnO₄⁻ + 5H₂O₂ + 6H⁺ → 2Mn²⁺ + 5O₂ + 8H₂O

✅ Correct Colours

  • From: colourless
  • To: (pale) pink

🧠 What Is in the Flask?

The acidified H₂O₂ solution is in the conical flask (colourless). The purple KMnO₄ solution is added from the burette.

During titration, MnO₄⁻ is immediately reduced to colourless Mn²⁺ . At the end-point, one drop of excess MnO₄⁻ imparts a persistent pale pink tint.

❌ Examiner Trap

Do NOT write "purple", "dark pink", or "red" for the final colour. The end-point is reached at the first permanent trace of unreacted manganate(VII), which appears pale pink. Purple indicates significant over-titration.

Mark Scheme: Mark 1: From colourless. Mark 2: To (pale) pink. (Reject "purple" or "dark pink"). Allow 1 mark if colours are reversed.

Part (d): Multi-Step Quantitative Titration Calculation

Deducing whether H₂O₂ concentration satisfies WHO guidelines (5 Marks)

📐 Step-by-Step Calculation

1 Moles of KMnO₄ used in titration:
n(KMnO₄) = c × V = 0.0200 mol dm⁻³ × (15.60 / 1000) dm³ = 3.12 × 10⁻⁴ mol

2 Moles of H₂O₂ in 25.00 cm³ diluted sample:
From the balanced equation, ratio is 2 MnO₄⁻ : 5 H₂O₂ .
n(H₂O₂ in 25.0 cm³) = (3.12 × 10⁻⁴ ÷ 2) × 5 = 7.80 × 10⁻⁴ mol

3 Moles of H₂O₂ in full 250.0 cm³ volumetric flask:
The 25.00 cm³ sample was taken from 250.0 cm³ (dilution factor = 250 / 25.0 = 10).
n(H₂O₂ in 250 cm³) = 7.80 × 10⁻⁴ × 10 = 7.80 × 10⁻³ mol
This is the total amount of H₂O₂ in the original 10.00 cm³ sanitiser sample.

4 Mass of H₂O₂ in sanitiser:
Molar mass of H₂O₂ = (2 × 1.0) + (2 × 16.0) = 34.0 g mol⁻¹
Mass of H₂O₂ = 7.80 × 10⁻³ mol × 34.0 g mol⁻¹ = 0.2652 g

5 Mass of hand sanitiser and percentage by mass:
Mass of sanitiser = density × volume = 0.82 g cm⁻³ × 10.00 cm³ = 8.20 g
% by mass = (0.2652 g / 8.20 g) × 100 = 3.23% (or 3.2%)
Conclusion: 3.23% lies between 3.0% and 6.0%, so it is within World Health Organization guidelines.

❌ Common Calculation Traps

  • Inverting the mole ratio: Multiplying by 2/5 instead of 5/2.
  • Forgetting the density: Dividing by 10.0 g instead of 8.20 g. Volume must be converted to mass using density ( mass = d × V ).
  • Missing conclusion: Forgetting to state whether the calculated percentage is within the 3.0%–6.0% range.

🧠 Significant Figures

Density is provided to 2 SF (0.82 g cm⁻³), so the final percentage is best quoted to 2 or 3 SF ( 3.2% or 3.23% ). The mark scheme accepts TE (transferred errors) throughout, but strictly penalises answers rounded to 1 SF.

Mark Scheme: Mark 1: Moles of KMnO₄ (3.12 × 10⁻⁴). Mark 2: Moles of H₂O₂ in 25 cm³ (7.80 × 10⁻⁴). Mark 3: Moles of H₂O₂ in 250 cm³ (7.80 × 10⁻³). Mark 4: Mass of H₂O₂ (0.2652 g). Mark 5: % by mass (3.23%) and correct deduction.

Part (e): Percentage Uncertainties

Pipette vs. Burette percentage error calculation (2 Marks)

📐 Pipette Uncertainty (1 Mark)

A volumetric pipette is a single-delivery piece of apparatus (involves one reading):

% uncertainty = (0.04 / 25.00) × 100 = (±) 0.16%

📐 Burette Uncertainty (1 Mark)

A titre is obtained by taking two readings (initial and final), so total uncertainty is doubled:

% uncertainty = (2 × 0.05 / 15.55) × 100 = (±) 0.643% (or 0.64%)

❌ Critical Distinction: 1 vs 2 Readings

A very common lost mark occurs in the burette calculation when students forget to multiply by 2:

(0.05 / 15.55) × 100 = 0.32% ❌ Incorrect! Titres always require two readings.

(2 × 0.05 / 15.55) × 100 = 0.64% ✅ Correct!

Mark Scheme: Mark 1: Pipette % uncertainty = 0.16%. Mark 2: Burette % uncertainty = 0.643% or 0.64% (using 2 × 0.05). Penalise 1 SF once only.

Topics

Physical Chemistry · Topic 3: Redox I · Topic 5: Formulae, Equations and Amounts of Substance

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.